Graphical methods - dynamics problem.

A car and a van are at rest on a straight horizontal road, with the van 25m in front of the car. At time t=0 seconds, the van moves off with an acceleration of 1.5 m/s². At time t=5 seconds, the car moves off in the same direction along the road with an acceleration of 2 m/s². use the sketches (I did a computer sketch and it can be found here) of the velocity/time graphs of the car and van, on the same set of axes, to calculate the time at which their velocities are the same and state this velocity. Also, from the sketch, determine the distance between the van and car at this time.

I equated the two equations to get the velocity being the same at t = 20, and substituted this in the van's equation to get that velocity to be v= 30. My main problem is with finding the distance between the van and the car at that time. The answer tell me it's 100m, but I've no idea what the method is. Please help! Thanks.

The point of the exercise was to realise that the area under a certain line on a velocity/time graph equals the distance travelled. I then assumed that if I found the distance the van had travelled, and subtracted from it the distance the car had travelled, I would have my answer. But the answer I got that way did not equal the answer in the book...

The answer's meant to be 100...but again, I'm not meant to be using integration. I'm meant to be using the area under each graph as the distance travelled...so looking at the graph, before reaching t=20, the van travelled:

the area under the blue line between t=0 and t=20: 1/2 x 20 x 30 = 300m

and the car travelled:

the area under the green line between t=0 and t=20: 1/2 x 15 x 30 = 225m

reread the problem carefully. It states that the van starts from a position 25 ahead of the car. You have taken into account the offset by virtue of the headstart, but not the physical offset that has nothing to do with time.